3. Rebecca puts her backpack on a fan cart and turns the fan on. She then measures the speed each second afterward. At 10 seconds the cart has a speed of 20 cm
per second. How fast will the cart be moving at 12 seconds?
O A 18 cm per second
OB 20 cm per second
OC.22 cm per secnod
01.24 cm per second

50 points please help :)

There are 2500 grams of sugar (sucrose) in 5 liters of sugar water with a concentration of 0.5 grams per liter of solution.

To calculate the grams of sugar in 5 liters of sugar water with a concentration of 0.5g per liter, we need to use the formula:

mass = concentration x volume

First, we convert 5 liters to milliliters by multiplying by 1000:

5 L x 1000 mL/L = 5000 mL

Then, we plug in the values:

mass = 0.5 g/L x 5000 mL

mass = 2500 g

Therefore, there are 2500 grams of sugar (sucrose) in 5 liters of sugar water with a concentration of 0.5g per liter of solution.

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The concentration of the ions in the solution cannot be determined from the information provided. To determine the concentration, additional information, such as the initial concentrations of the reactants and any reaction conditions, would be needed.

Generally, Silver chromate (Ag2CrO4) is a chemical compound made of silver, chromium, and oxygen ions. It is an insoluble substance that forms a red precipitate in aqueous solution.

From the information that has been supplied, it is not possible to estimate the amount of ions that are present in the solution.

Additional information, such as the beginning concentrations of the reactants and any circumstances that may be present throughout the reaction, would be required in order to calculate the concentration.

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The correct answer is (c) Two of these choices are correct. The rate constant (k) of a chemical reaction is affected by both the concentration of the reactants and the temperature. The rate constant does not depend on the nature of the reactants or the order of the reaction.

The concentration of the reactants affects the rate constant through the rate law equation, which relates the rate of the reaction to the concentrations of the reactants. For example, for a first-order reaction, the rate law equation is:

rate = k[A]

where [A] is the concentration of the reactant A. As the concentration of A increases, the rate constant also increases.

The temperature affects the rate constant through the Arrhenius equation, which relates the rate constant to the activation energy and the temperature. The Arrhenius equation is:

k = Ae^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. As the temperature increases, the rate constant also increases exponentially.

The nature of the reactants and the order of the reaction do not affect the rate constant. The nature of the reactants affects the rate of the reaction, but not the rate constant. The order of the reaction affects the rate law equation, but not the rate constant.

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Answer:

I feel like the answer is A) true

Explanation:

you create a model to help your audience understand your complex idea. There have been models made to represent ideas that we cannot see.

The statement "Isotopes with an even number of both protons and neutrons are generally stable" is TRUE.

Isotopes with an even number of protons and neutrons are known as isotopes. A single element can have a different number of neutrons; such elements are known as isotopes. A nucleus containing a different number of neutrons will have a different mass number than the element's standard atomic number since the mass number is equal to the number of neutrons plus the number of protons in the nucleus.

There are 275 isotopes recognized for 81 stable elements, while there are 50 isotopes for elements that are radioactive and occur naturally.

Isotopes with an even number of both protons and neutrons tend to be more stable, as the strong nuclear force and the electromagnetic force in a nucleus cancel out, causing a more stable nucleus. The vast majority of stable elements have a nearly equal number of neutrons and protons, though there are some exceptions, such as beryllium-8 and helium-3.

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The aPTT result, in this case, is 55 seconds. The nurse should follow the protocol. If the aPTT is within the recommended range, the infusion dose will remain the same. Heparin is an anticoagulant drug that can prevent blood clots from forming.

aPTT is the laboratory test used to monitor heparin therapy. It is important to remember that heparin can cause bleeding if given in higher doses. For that reason, it is vital to adjust the dose of the heparin if the aPTT is outside the recommended range.Therefore, in this scenario, since the aPTT is within the recommended range, the nurse should adjust nothing. As a result, the infusion dose will remain the same as it was before. After the first aPTT is completed, the nurse should repeat it every six hours, as directed by the protocol. In conclusion, in this scenario, the nurse should do nothing since the aPTT is within the recommended range. The nurse should follow the protocol by rechecking aPTT every six hours.

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Taking into account the definition of molarity, the molarity of the solution is 0.186\(\frac{moles}{liter}\).

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

\(Molarity=\frac{number of moles}{volume}\)

Molarity is expressed in units \(\frac{moles}{liter}\).

In this case, you know:

Replacing in the definition of molarity:

\(Molarity=\frac{3x10^{-3} moles}{0.0161 L}\)

Solving:

molarity= 0.186\(\frac{moles}{liter}\)

Finally, the molarity of the solution is 0.186\(\frac{moles}{liter}\).

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α-anomer of d-glucose less likely to form than the β-anomer because the β-anomer undergoes less electron repulsion.

The hydroxyl group on the anomeric carbon of the β-anomer is equatorial, thereby creating less steric hindrance than the α-anomer, which has the hydroxyl group of the anomeric carbon in axial position.

(R)-D- and (S)-L-glyceraldehyde serve as reference points for the assignment of relative configuration to all other carbohydrates such as glucose. The reference point is the chiral center farthest from the carbonyl group. A D-carbohydrate has the same configuration at its farthest chiral center as D-glyceraldehyde (its -OH is on the right when written as a Fischer projection); an L-carbohydrate has the same configuration at its farthest chiral center as L-glyceraldehyde (its -OH is on the left). D-Glucose is the stereoisomer found in living systems. The other chiral centers in D-glucose characterize it as a "glucose," not a different sugar like mannose or galactose (other carbohydrate stereoisomers), but only the last chiral centre is important for distinguishing a carbohydrate as being D or L. Before continuing, you might want to make sure you understand this definition because it's difficult for most students to grasp.

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An aspect of the kinetic-molecular theory and can be used to explain the behavior of plasmas is particle exchange energy through elastic collision.

In physics, an elastic collision is an encounter (collision) between the two bodies in which the total kinetic energy of the two bodies remains the same. In an ideal, perfectly elastic collision, there is no net conversion of the kinetic energy into other forms such as heat, noise, or potential energy.

During collision of small objects, kinetic energy is first converted to potential energy associated with a repulsive or attractive force between the particles (when the particles move against this force, i.e. angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. angle between force and the relative velocity is acute).

Collisions of the atoms are elastic, for example Rutherford backscattering.

A useful special case of the elastic collision is when the two bodies have equal mass, in which case they will simply exchange their momenta.

The molecules—as distinct from atoms—of gas or liquid rarely experience perfectly elastic collisions because kinetic energy is exchanged between the molecules’ translational motion and their internal degrees of freedom with each collision. At any instant, half collisions are, to a varying extent, inelastic collisions (the pair possesses less kinetic energy in their translational motions after the collision than before), and half could be described as “super-elastic” (possessing the more kinetic energy after the collision than before). Averaged across the entire sample, the molecular collisions can be regarded as essentially elastic as long as Planck's law forbids energy from being carried away by black-body photons.

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Serine proteases are enzymes that cleave peptide bonds in proteins. Serine serves as the nucleophilic amino corrosive at the dynamic site of the enzyme. They are found ubiquitously in both prokaryotes and eukaryotes.

The catalytic triad gives a paradigm for the auxiliary and chemical highlights of enzymes that permit them to encourage a difficult response. The reaction in this case is the hydrolysis of a peptide bond, which - in spite of the fact that thermodynamically favorable - is kinetically blocked off beneath ordinary physiological conditions. The course of chemicals known as serine proteases - of which chymotrypsin may be a part - have over a long time been exceptionally completely characterized by a wide assortment of biochemical and auxiliary strategies. Buildups of chymotrypsin critical to its catalytic work were, to begin with, distinguished utilizing methods of protein chemistry such as partiality names in combination with a reasonable measure for the enzyme's action. In this way, the structures of chymotrypsin and other serine proteases uncovered that the dynamic destinations of these chemicals shared a specific stereochemical course of action of residue, together with a histidine (His) and an aspartate (Asp) buildup. The conjures uncommon properties of the catalytic set of three to clarify the huge improvement of the peptide hydrolysis response watched for the serine proteases.

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The activity of the enzyme lysine decarboxylase is to remove the carboxyl functional group from the amino acid lysine.

This process is known as decarboxylation and results in the formation of the molecule cadaverine. Lysine decarboxylase is an important enzyme involved in the breakdown of proteins and amino acids in living organisms. Decarboxylation is a common biochemical reaction that occurs in many different metabolic pathways.

This process is important for the synthesis of other important molecules, such as polyamines, which are involved in cell growth and differentiation. Lysine decarboxylase is found in a variety of organisms, including bacteria, fungi, and plants. It is particularly important in bacteria, where it is involved in the production of biogenic amines, such as putrescine and cadaverine.

Overall, the activity of the enzyme lysine decarboxylase plays a crucial role in many different biological processes, from the breakdown of proteins to the production of important signaling molecules. Understanding the function and regulation of this enzyme is therefore of great importance for our understanding of basic biology and for the development of new therapeutics and treatments.

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If an electric discharge produce 300 L of ozone(O₃), the liter of oxygen(O₂) are required is 450 L.

The reaction is as follows :

3O₂  --->  2O₃

The volume of the Ozone = 300 L

The mole ratio is given as :

3 moles of the oxygen that is O₂  will produce the 2 moles of the ozone that is O₃.

The number of the moles is as follows :

The number of moles = mass / molar mass

The 300 L of the ozone will required the oxygen is as :

To produce 300 L of ozone  = ( 3/2 ) × 300 L of the oxygen required

                                               = 1.5 × 300 L

                                               = 450 L

Thus, the 450 L of the oxygen is required to produce the 300 mL of the ozone .

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Long-term heat loss to the calorimeter is a problem, especially for slow-moving processes.

The water iscontained in a tank whose thermal capacity and weight were predete partially rmined before the experiment began. Calculating heat and flow rates involves measuring the rise in water temperature and volume over a predetermined period of time.

In contrast, if the reaction absorbs heat, heat is transmitted from the calorimeter to the system, lowering the calorimeter's temperature.

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Answer: An Ion

Explanation:

Ions with a positive charge are called cations. Ions with a negative charge are called anions.

Atoms of different element can not have same atomic number because only same type of atoms combine to form element. Atoms belonging to different element can have different atomic number. Therefore, the ore of aluminum is Bauxite.

Element generally consist of atoms or we can atoms combine to form element. Atoms of an element is always same, means all the properties of all atoms of one type of element is same.

Two or more than two atoms with different physical or chemical properties can not combine together to form an element.  Ore is a naturally occurring solid material from which a metal or valuable mineral can be extracted profitably. The ore of aluminum is Bauxite.

Therefore, the ore of aluminum is Bauxite.

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Answer:

1.238 years will it take

Explanation:

The isotope decay of an atom follows the equation:

Ln[A] = -kt + ln[A]₀

Where [A] could be taken as the amount of the isotope after t years

[A] = 1.00mg

t = ?

k is the decay constant; k = ln2 / Half-life

k = ln 2 / 2.86 years = 0.2424 years⁻¹

[A]₀ is the initial amount of the isotope = 1.35mg

Replacing:

ln1 = -0.2424years⁻¹*t + ln 1.35

0.2424years⁻¹*t = ln 1.35

t =

Answer:

Copper(cu) nd gold(au) will not react with acid at all

Explanation:

The electron will move to energy level

n =1 ..

We have, an electron in the n= 5 level of an Hydrogen atom emits a photon of wavelength 94.98nm...

Using the Rydberg formula ,

1/ λ = R ( 1/ n²ⱼ- 1/n²ᵢ)

where, lambda ----> wavelength of photon

nⱼ----> excited energy level

nᵢ---> initial energy level at which electron of hydrogen atom emits photon

R -----> Rydberg constant

photon are packet of energy.

we have give nᵢ = 5 , lambda = 94.98nm = 94.98×10⁻⁹m

R = constant = 1.097× 10⁷ m⁻¹

Substitute the values in formula we get ,

10⁹/ 94.98 = 1.097× 10⁷ ( 1/ n²ⱼ - 1/ 25) ⇒ 1/ n²ⱼ - 1/25 = 10^2/94.98×1.097 = 0.95975.68 ⇒ 1/n²ⱼ= 0.99975 ⇒ n²ⱼ= 1.00024 => nj = 1.00012

So, the electron moves from n= 5 energy level to to n= 1 energy level.

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First of all, Z is unknown. I hope it is a mistake.

        Now, it is given that the element X has nineteen electrons. This proves that X is actually Potassium.

        As per the periodic table, both Potassium and Lithium belongs to group 1 as their valency is 1 because of the presence of only one electron in the outermost shell of electrons i.e., they lose an electron during a chemical reaction to form a stable compound. Furthermore, both are metallic.

        Magnesium belongs to group 2 and hence its valency is two, which is different from potassium though it is metallic. Similiarly, bromine belongs to group 17 and gains one electron during a reaction in contrast to potassium.

( No internal links available for reference. For clarification, check the periodic table).

Answer:

Orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron, and all electrons in singly occupied orbitals must have the same spin state. Noble-gas configuration refers to an outer main energy level occupied, in most cases, by eight electrons.

Explanation:

I hope this helps you.

Answer:

Two electrons

Explanation:

Each individual orbital can hold a maximum of two electrons. Hope this helps :)

A radioactive substance decreases by 65% each hour. Find the hourly decay factor. The hourly decay factor is 0.35.

Chemicals in the class of radionuclides (also known as radioactive materials) have unstable atomic nuclei. They become stable by undergoing modifications in the nucleus (spontaneous fission, alpha particle emission, neutron conversion to protons, or the opposite).

A radioactive atom will naturally emit radiation in the form of energy or particles in order to transition into a more stable state. The difference between radioactive material and the radiation it emits must be made.

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Answer:

A food web is made up of all the food chains in a single ecosystem. Each living thing in an ecosystem belongs to many food chains. A food chain is a path that energy takes through a certain ecosystem. Trophic Levels. Organisms in food webs are grouped into categories called trophic levels.

Explanation:

I HOPE THIS HEPLS HAVE A WONDERFUL DAY!:)

Answer:

you didnt say if it was an animal or people food web so imma just try my best to explane tis is an animal food web. So it really is just showing was the animals eat and if they eat other animal basicly.

Explanation:

we write the empirical formula using the mole ratios as subscripts: Empirical formula of Calcocite = Cu2STherefore, the empirical formula for Calcocite is Cu2S.

Calcocite is a mineral composed of 79.9 mass % copper and 20.1 mass % sulfur. To determine the empirical formula for Calcocite, we must first determine the mole ratios of copper and sulfur in the mineral.

Here's how: Step-by-step explanation:

We assume a 100 g sample of Calcocite.

Then, the mass of copper in the sample is 79.9 g (79.9 mass % of 100 g) and the mass of sulfur in the sample is 20.1 g (20.1 mass % of 100 g).Next, we find the moles of copper and sulfur using their atomic masses:

Moles of copper = 79.9 g / 63.55 g/mol (atomic mass of copper)

= 1.257 moles Moles of sulfur

= 20.1 g / 32.06 g/mol (atomic mass of sulfur)

= 0.626 moles

Then, we divide each mole value by the smallest mole value to get the simplest whole-number ratio of moles:

Moles of copper / Smallest mole value

= 1.257 / 0.626

= 2.007 Moles of sulfur / Smallest mole value

= 0.626 / 0.626

= 1.000 (Note that we rounded off the mole ratio of copper to two decimal places, which may introduce some errors in our calculation.

However, we will use this rounded value to make our empirical formula easier to write.)

Finally, we write the empirical formula using the mole ratios as subscripts: Empirical formula of Calcocite = Cu2STherefore, the empirical formula for Calcocite is Cu2S.

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Benedict's test should be done to prove it.

Glucose is the main sugar which is mainly found in our blood. it is present in carbohydrate products. People with fructose intolerance should avoid peas because it contain a lot of fructose

Benedict's test is a test where chemicals are used to check the presence of sugar in a particular substance. The sugar that is present in the substance will react with the reducing sugar present in the substance on heating. the positive result is shown by the colour change from clear blue to brick red precipitate.

Pea contains a lot of fructose and it can be proved by doing benedict test.

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The system's effort in the process known as isothermal can be determined using the equation. W = 2.303RT log 10 (V 2 / V 1), where V is the volume in two distinct states that are both at temperature T, and R is the constant of all gases.

Although a refrigerator's mechanism goes through a number of changes, the internal temperature never varies. Here, warmth energy is released and spread throughout the nearby area. The heat pump is an additional instance of an isothermal mechanism.

Isothermal compression is a thermodynamic mechanism that lowers the volume or raises the pressure in a system with a constant temperature. Thermal equilibrium is preserved by the procedure.

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After a workout, most people show signs of strength, warmth, contentment, fulfillment, success, and accomplishment.

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The ∆G° for the dissolution of AgCl solid is 55.7 kJ/mol.

The gibbs free energy is termed as thermodynamic quantity which is equal to the enthalpy (of a system or process) subtracted by the product of the entropy and the absolute temperature.

The gibbs free energy of the reaction is given by:

∆G°(rxn) =sum of ([∆G°] products)- sum of ( [∆ G°] reactants)

Reaction can be given as:

AgCl(S) » Ag+ (aq) + Cl- (aq)

Gibbs free energy of silver ions = ∆G°f, (Ag+) = 77.1kJ/mol

Gibbs free energy of formation of chloride ions = ∆G°f, (Cl-)= -131. 2kJ/mol

Gibbs free energy of formation of silver chloride solid = ∆ G°f, (AgCl) = - 109.8kJ/mol

The gibbs free energy of the reaction of dissolution of AgCl) :∆ G°rxn:

∆G°rxn = (77.1kJ/mol) – (- 109.8kJ/mol)

55.7kJ/mol

Thus, we found that the ∆G° for the dissolution of AgCL solid is 55.7 kJ/mol.

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The equilibrium temperature when a 235 g block of copper at 255 ∘c is placed in a 135 g aluminum calorimeter cup containing 825 g of water at 13.0 ∘c is 22.9 ∘C.

The amount of heat lost by the copper block Q gained is the amount of heat gained by the water and the aluminum cup. For Copper Q = mCpΔTFor water and aluminum cup Q = (mCpΔT)water + (mCpΔT)aluminum. We need to find out the final temperature of the copper block, water, and aluminum cup system.

Now, we can substitute the values in the equation to find out the final temperature.(235 g)(0.385 J/g∙K)(255 ∘C - Tfinal) = (135 g)(0.903 J/g∙K) (Tfinal - 13.0 ∘C) + (825 g)(4.184 J/g∙K) (Tfinal - 13.0 ∘C)(90497.5 J/K) - (90497.5 J/K) Tfinal = (1248.9 J/K) Tfinal - 17138.4 J/K + (3451236 J/K) Tfinal - (450828 J/K)(3450237.5 J/K) T final = 4540647.5 J/KTfinal = 1.316 K or -271.834 ∘CThe final temperature of the system cannot be negative.

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